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I haven't done linear algebra for a while, so, please bear with me. Let $d\in\mathbb N$ and $U$ be a $(d-1)$-dimensional subspace of $\mathbb R^d$. It's easy to show that there is some $\eta$ in the algebraic dual space of $\mathbb R^d$ with $$U=\ker\eta\;.\tag1$$ Now, let $P:\mathbb R^d\to\mathbb R$ be a polynomial of degree $d$ with $$\left.P\right|_U=0\;.\tag2$$

I want to show that there is a polynomial $Q:\mathbb R^d\to\mathbb R$ of degree $d-1$ with $$P=\eta Q\;.\tag3$$

I struggle with the question why we can assume that $$\eta(x)=x_d\;\;\;\text{for all }x\in\mathbb R^d\tag3$$ and $$U=\mathbb R^{d-1}\times\left\{0\right\}\tag4\;.$$ Actually, it's intuitively clear to me, but I don't remind how we need to argue rigorously. Sure, we somehow need to find a function $T:\mathbb R^d\to\mathbb R^d$ (which rotates and translates $U$) such that $$T(U)=\mathbb R^{d-1}\times\left\{0\right\}\tag5\;.$$ How can we do that and why is the assumption of $(3)$ and $(4)$ legitimate?

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    It looks like it is a question of choice of a basis: choose the basis $B$ such that the first $d-1$ vectors form a basis of $U$. Then $\eta$ is really the projection onto the last coordinate. Choosing different basis corresponds to a transformation $T$ as mentioned. – 2017-02-13
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    @pepa.dvorak If we choose a basis $(u_1,\ldots,u_{d-1})$ of $U$ and denote the standard basis of $\mathbb R^d$ by $(e_1,\ldots,e_d)$, we only know that there is *some* $i\in\left\{1,\ldots,d\right\}$ such that $(u_1,\ldots,u_{d-1},e_i)$ is a basis of $\mathbb R^d$. So, $\eta$ is the projection onto the $i$-th coordinate. (Note that in $\mathbb R^d$ we usually always understand "coordinate" with respect to the standard basis and that's how $x_d$ in $(3)$ has to be understood; sure, $\eta$ is the projection onto the last coordinate with respect to $(u_1,\ldots,u_{d-1},e_i)$.) – 2017-02-14
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    @pepa.dvorak And I need to correct myself, $\eta$ doesn't even need to be the projection onto the $i$-th coordinate, we only know that $$\eta\left(\sum_{j=1}^{d-1}\alpha_ju_j+\beta e_i\right)=\beta\eta(e_i)\;.$$ – 2017-02-14
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    (3) (the first one, in the box) is invariant under a linear change of coordinates, so we just choose coordinates $x$ that make (3), (4) happen and at the end go back to the original coordinates $y=Tx$. – 2017-02-17
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    @ChristianRemling Could you please prove an explicit explanation as an answer? What kind of "linear change of coordinates" exactly, why are we allowed to do that and how can we "go back to the original coordinates"? That are the parts that I don't understand. – 2017-02-18
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    Choose coordinates $x$ so that (3), (4) hold. Then $P(x)=x_d Q(x)$, so $P(y)=(Sy)_d Q(Sy)$, which is of the desired form since $y\mapsto (Sy)_d$ can be our $\eta$. – 2017-02-18

2 Answers 2

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Some basic facts about polynomials and linear transformations on $\mathbb R^d:$

Lemma 1: If $P$ is a polynomial on $\mathbb R^d$ of degree $n$ and $T:\mathbb R^d \to \mathbb R^d$ is a nonsingular linear transformation, then $P\circ T$ is a polynomial on $\mathbb R^d$ of degree $n.$

Lemma 2: If $P(x_1,\dots,x_d)$ is a polynomial on $\mathbb R^d$ of degree $n>0$ and $P=0$ on $\mathbb R^{d-1}\times \{0\},$ then there exists a polynomial $Q$ of degree $n-1$ such that $P(x) = x_dQ(x)$ for all $x\in \mathbb R^d.$

Lemma 3: If $U$ is a $(d-1)$ dimensional subspace of $\mathbb R^d,$ and $\eta_1, \eta_2:\mathbb R^d \to \mathbb R$ are linear maps with $\ker\eta_1= \ker \eta_2=U,$ then there is a nonzero constant $c\in \mathbb R$ such that $\eta_2=c \eta_1$ on $\mathbb R^d.$

Back to the problem, where we have $P=0$ on $U = \ker \eta,$ with $P$ a polynomial of degree $n>0.$ (I'm not sure why you have $n=d$ in the problem; the result is true for arbitrary $n.$)

Choose a basis $u_1,\dots,u_{d-1}$ for $U,$ and extend it to a basis $u_1,\dots,u_{d-1},u_d$ of $\mathbb R^d.$ Let $e_1,\dots ,e_d$ be the standard basis for $\mathbb R^d.$ Then there is a unique nonsingular linear $T:\mathbb R^d \to \mathbb R^d$ such that $T(e_k) = v_k$ for all $k.$ We have $T$ mapping $\mathbb R^{d-1}\times \{0\}$ isomorphically onto $U,$ with $T^{-1}$ "doing the same" coming back.

By Lemma 1, $P\circ T$ is a polynomial of degree $n>0$ that vanishes on $\mathbb R^{d-1}\times \{0\}.$ By Lemma 2, $P\circ T(x) = x_dQ(x)$ for some polynomial $Q$ of degree $n-1.$ Let $\Pi(x) = x_d, x\in \mathbb R^d.$ With this notation we have $P\circ T = \Pi\cdot Q$ on $\mathbb R^d.$ Therefore $P = (\Pi\circ T^{-1})\cdot (Q \circ T^{-1}).$ Note that $Q \circ T^{-1}$ is a polynomial of degee $n-1$ by Lemma 2.

Because $T^{-1}$ maps $U$ to $\mathbb R^{d-1}\times \{0\},$ and $\Pi =0$ on this last set, we have $\Pi\circ T^{-1} = c\eta$ by Lemma 3. Thus $P = (c\eta)\cdot (Q \circ T^{-1}) = \eta\cdot (cQ \circ T^{-1}).$ This shows $P$ has the desired form and we're done.

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As pointed out by @pepa.dvorak in the comment, the problem can probably be solved by an appropriate basis change. My solution is more complicated and based on a more general approach.

Since the ideals $(\eta)$ and $(\eta,P)$ generate the same affine variety, their radical ideals coincide. In particular, we get $P^k=\eta Q$ with some positive $k$, which means that $\eta$ divides $P$.